How to find the inverse of a function

In summary: Awesome! Thank you soooo much. I spent about 4-5 hours trying to figure this problem out myself and I wouldn't have gotten through it by myself.Logarithms are the inverse of the finish of exponentiation, by definition. Hopefully you remember your first lesson about them. Well hopefully they told you that then.
  • #1
Kupkake303
4
0
moved into h/w help, so template is missing
T(t) = Ts+(98.6 – Ts)e-kt

rewrite in the form t=g-1(T)

In trying to understand how to find the inverse of this but am having a hard time, please advise.

Thanks,
Kupkake303
 
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  • #2
This is just a matter of trying to get [itex]t[/itex] by itself on one side of an equality. You initially have: [itex]T = T_s + (98.6 - T_s) e^{-kt}[/itex]. To simplify the expression, let [itex]U = e^{-kt}[/itex]. Then the equation is: [itex]T = T_s + (98.6 - T_s) U[/itex]. So solve for [itex]U[/itex]. Then solve for [itex]t[/itex] in terms of this value of [itex]U[/itex].
 
  • #3
stevendaryl said:
This is just a matter of trying to get [itex]t[/itex] by itself on one side of an equality. You initially have: [itex]T = T_s + (98.6 - T_s) e^{-kt}[/itex]. To simplify the expression, let [itex]U = e^{-kt}[/itex]. Then the equation is: [itex]T = T_s + (98.6 - T_s) U[/itex]. So solve for [itex]U[/itex]. Then solve for [itex]t[/itex] in terms of this value of [itex]U[/itex].
So I would put it as 0=Ts+(98.6-Ts)U ??

Doing that would give me -Ts/(98.6-Ts)=U

Then putting the e-kt back in you get -Ts/(98.6-Ts)=e-kt

So then I would then need to get that by it's self so the t is by itself but I can't remember how to do this. Please advise. Thank you
 
  • #4
Kupkake303 said:
So I would put it as 0=Ts+(98.6-Ts)U ??

No, [itex]T = T_s + (98.6 - T_s)U[/itex]

So [itex]U = (T-T_s)/(98.6 - T_s)[/itex]
 
  • #5
stevendaryl said:
No, [itex]T = T_s + (98.6 - T_s)U[/itex]

So [itex]U = (T-T_s)/(98.6 - T_s)[/itex]
Right, because the T wouldn't just disappear.

So next I would change the U back to a e-kt for e-kt=(T-Ts)/(98.6-Ts)

from there would I try to figure out the e-kt ?
 
  • #6
Kupkake303 said:
Right, because the T wouldn't just disappear.

So next I would change the U back to a e-kt for e-kt=(T-Ts)/(98.6-Ts)

from there would I try to figure out the e-kt ?

You want to take the natural log of both sides of the equation.
 
  • #7
stevendaryl said:
You want to take the natural log of both sides of the equation.

okay so I would
e-kt=(T-Ts)/(98.6-Ts)

lne-kt=ln(T-Ts)/(98.6-Ts)

The ln would cancel out the e

-kt=ln(T-Ts)/(98.6-Ts)

then divide by -k? to get

t=(1/-k)ln(T-Ts)/(98.6-Ts)...?

so this would be my answer for finding the inverse ..?
 
  • #8
Kupkake303 said:
okay so I would
e-kt=(T-Ts)/(98.6-Ts)

lne-kt=ln(T-Ts)/(98.6-Ts)

The ln would cancel out the e

-kt=ln(T-Ts)/(98.6-Ts)

then divide by -k? to get

t=(1/-k)ln(T-Ts)/(98.6-Ts)...?

so this would be my answer for finding the inverse ..?

Yes, except that [itex]-ln(A/B) = ln(B/A)[/itex], so you can get rid of the [itex]-[/itex] sign to get:

[itex]t = \frac{1}{k} ln((98.6 - T_s)/(T-T_s))[/itex]
 
  • #9
stevendaryl said:
Yes, except that [itex]-ln(A/B) = ln(B/A)[/itex], so you can get rid of the [itex]-[/itex] sign to get:

[itex]t = \frac{1}{k} ln((98.6 - T_s)/(T-T_s))[/itex]

Awesome! Thank you soooo much. I spent about 4-5 hours trying to figure this problem out myself and I wouldn't have gotten through it by myself.
 
  • #10
Logarithms are the inverse of the finish of exponentiation, by definition. Hopefully you remember your first lesson about them. Well hopefully they told you that then.
 

Related to How to find the inverse of a function

1. What is an inverse function?

An inverse function is a mathematical operation that "undoes" or reverses the effects of another function. It essentially switches the input and output values of the original function.

2. How do you find the inverse of a function?

To find the inverse of a function, you must follow these steps:1. Start with the original function, f(x).2. Replace f(x) with y.3. Switch the x and y variables in the function.4. Solve the new equation for y.5. Replace y with f^(-1)(x), the inverse function.

3. What is the notation for an inverse function?

The notation for an inverse function is f^(-1)(x), which reads as "f inverse of x". It is important to note that this does NOT mean 1/f(x), as it may appear.

4. How do you graph an inverse function?

To graph an inverse function, you can use the graph of the original function and reflect it over the line y=x. This means that any points on the original function with coordinates (x,y) will have coordinates (y,x) on the inverse function.

5. What is the horizontal line test and how does it relate to finding the inverse of a function?

The horizontal line test is a method used to determine if a function has an inverse. If a horizontal line intersects the graph of the function at more than one point, then the function does not have an inverse. This test is important because a function must be one-to-one (have a unique output for every input) in order for it to have an inverse.

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